If f(x) is an even function and satisfies the relation  $x^2\cdot f\left(x\right)-2\cdot f\left(\frac{1}{x}\right)=g\left(x\right)$ where g(x) is an odd function, then the value of f(5) is

$x^{2}f\left(x\right)-2\cdot f\left(\frac{1}{x}\right)=g\left(x\right)\Rightarrow$ replacing ‘x’ by  $\frac{1}{x}$, then  $\frac{1}{x^2}\cdot f\left(\frac{1}{x}\right)-2f\left(x\right)=g\left(\frac{1}{x}\right)$

or  $2f\left(\frac{1}{x}\right)-4x^{2}f\left(x\right)=2x^2\cdot g\left(\frac{1}{x}\right)\mathrm{...........}\left(2\right)$

(1) + (2)   $\Rightarrow f\left(x\right)=-\left(\frac{g\left(x\right)+2x^2\cdot g\left(\frac{1}{x}\right)}{3x^2}\right)$

$f(-x)=-\left(\frac{g(-x)+2x^2\cdot g(-\frac{1}{x})}{3x^2}\right)$

$f\left(x\right) = - f\left(x\right)$

( $\therefore$ g(x) is an odd and f(x) is an even function) 

$f\left(x\right)=0\Rightarrow f\left(5\right)=0$